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The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two jars kept separately at temperatures of $27\,^oC$ and $127\,^oC$ respectively is
$\sqrt {\frac{1}{2}} $
$\sqrt {\frac{8}{3}} $
$\frac{4}{3}$
$\frac{3}{4}$
Solution
$\frac{1}{2} m v^{2}=\frac{3}{2} k T$
where $\mathrm{k}$ is the Boltzmann constant
$\therefore v=\sqrt{\frac{3 k T}{m}}$
Now the de-Broglie wavelength is given by
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{m}} \sqrt{\frac{\mathrm{m}}{3 \mathrm{kT}}}$
$\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
$\therefore \frac{{{\lambda _{\rm{H}}}}}{{{\lambda _{{\rm{He}}}}}} = \sqrt {\frac{{{{\rm{m}}_{{\rm{He}}}}}}{{{{\rm{m}}_{\rm{H}}}}} \times \frac{{{{\rm{T}}_{{\rm{He}}}}}}{{{{\rm{m}}_{\rm{H}}}}}} $
$\sqrt {\frac{4}{2} \times \frac{{127 + 273}}{{27 + 273}}} $
$ = \sqrt {\frac{{4 \times 400}}{{2 \times 300}}} = \sqrt {\frac{8}{3}} $
